Question

A solid cylinder is rolling without sliding. What fraction of its total kinetic energy is associated to rotational motion ?

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

$\frac{\text{Rotational KE}}{\text{Total KE}}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}I{\omega }^2+\frac{1}{2}m{V}^2}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}m{V}^2[1+\frac{1}{2}]}=\frac{\frac{1}{2}m{R}^2\times \frac{V^2}{R^2}}{m{V}^2\times \frac{3}{2}}=\frac{1}{3}$